Solve-2sin-2x-3-sin-x-cos-x-2-0-

Question Number 1870 by alib last updated on 18/Oct/15

S o l v e

2 s i n 2 x - 3 (s i n x + c o s x) + 2 = 0

Answered by 112358 last updated on 19/Oct/15

2 s i n 2 x - 3 (s i n x + c o s x) + 2 = 0

4 s i n x c o s x - 3 s i n x - 3 c o s x + 2 = 0

2 {s i n}^{2} x + 4 s i n x c o s x + 2 {c o s}^{2} x = 3 (s i n x + c o s x)

2 {(s i n x + c o s x)}^{2} - 3 (s i n x + c o s x) = 0

(s i n x + c o s x) (2 s i n x + 2 c o s x - 3) = 0

W e o b t a i n t h a t

s i n x + c o s x = 0 o r 2 (s i n x + c o s x) - 3 = 0

(1) s i n x + c o s x = 0

∵ s i n x = t a n x c o s x

\Rightarrow t a n x c o s x + c o s x = 0

c o s x (t a n x + 1) = 0

\Rightarrow c o s x = 0 \Leftrightarrow x = 2 n π \pm \frac{π}{2} n \in Z

o r t a n x = - 1 \Leftrightarrow x = m π - \frac{π}{4} m \in Z

C h e c k i n g t h e s o l u t i o n x = 0.5 π

2 s i n π - 3 (s i n 0.5 π) - 3 c o s 0.5 π + 2 = - 1 \neq 0

∴ x \neq 2 n π \pm 0.5 π

D e f i n e S_{1} a s s o l u t i o n s e t 1 .

∴ S_{1} = {x \in R, m \in N ∣ x = m π - \frac{π}{4}}

(2) 2 (s i n x + c o s x) - 3 = 0

I n h a r m o n i c f o r m

s i n x + c o s x = \sqrt{2} s i n (x + \frac{π}{4})

∴ 2 \sqrt{2} s i n (x + \frac{π}{4}) = 3

s i n (x + \frac{π}{4}) = \frac{3}{2 \sqrt{2}}

B u t, \frac{3}{2} = 1.5 > \sqrt{2} = 1.41 \dots \Rightarrow \frac{3}{2 \sqrt{2}} > 1 .

S i n c e ∣ s i n p ∣⩽ 1 \forall p \in R, \Rightarrow s i n (x + \frac{π}{4}) = \frac{3}{2 \sqrt{2}}

h a s n o r e a l s o l u t i o n .

T h e o n l y s e t o f r e a l s o l u t i o n s i s

S_{1} = {x \in R ∣ x = m π - \frac{π}{4}, m \in N}

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